Problem: What is the extraneous solution to these equations? $\dfrac{x^2 + 5x}{x + 3} = \dfrac{-6}{x + 3}$
Solution: Multiply both sides by $x + 3$ $ \dfrac{x^2 + 5x}{x + 3} (x + 3) = \dfrac{-6}{x + 3} (x + 3)$ $ x^2 + 5x = -6$ Subtract $-6$ from both sides: $ x^2 + 5x - (-6) = -6 - (-6)$ $ x^2 + 5x + 6 = 0$ Factor the expression: $ (x + 3)(x + 2) = 0$ Therefore $x = -3$ or $x = -2$ At $x = -3$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -3$, it is an extraneous solution.